Question

a large cruise ship of mass 6.80×10^7 kg has a speed of 12.2 m/s at some instant. (a) what is the ships kinetic energy at this time? (b) how much work is required to stop it? (give the work done on the ship. include the sign of the value in your answer.) (c) what is the magnitude of the constant force required to stop it as it undergoes a displacement of 2.80 km?

Explanation:

Step1: Calculate kinetic energy

The formula for kinetic energy is $K = \frac{1}{2}mv^{2}$, where $m = 6.80\times 10^{7}\text{ kg}$ and $v = 12.2\text{ m/s}$.
$K=\frac{1}{2}\times(6.80\times 10^{7})\times(12.2)^{2}$
$K = \frac{1}{2}\times6.80\times 10^{7}\times148.84$
$K= 5.06096\times 10^{9}\text{ J}$

Step2: Determine work - energy theorem for stopping the ship

According to the work - energy theorem, the work done $W$ on an object to change its kinetic energy is $W=\Delta K$. To stop the ship, the final kinetic energy $K_f = 0$ and the initial kinetic energy $K_i$ is the value calculated above. So $W=K_f - K_i=0 - 5.06096\times 10^{9}\text{ J}=- 5.06096\times 10^{9}\text{ J}$

Step3: Calculate the force using work - displacement relation

The work done by a constant force $F$ over a displacement $d$ is $W = Fd$. We know $W=-5.06096\times 10^{9}\text{ J}$ and $d = 2.80\times10^{3}\text{ m}$. Rearranging for $F$, we get $F=\frac{W}{d}$.
$F=\frac{-5.06096\times 10^{9}}{2.80\times 10^{3}}$
$F=-1.8074857\times 10^{6}\text{ N}\approx - 1.81\times 10^{6}\text{ N}$, and the magnitude of the force is $|F| = 1.81\times 10^{6}\text{ N}$

Answer:

(a) $5.06\times 10^{9}\text{ J}$
(b) $- 5.06\times 10^{9}\text{ J}$
(c) $1.81\times 10^{6}\text{ N}$